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[Ans : -15]vi) In a workshop a worker measures the lengthof a steel plate with a Vernier callipershaving a least count 0.01 cm. Four suchmeasurements of the length yielded thefollowing values: 3.11 cm, 3.13 cm, 3.14cm, 3.14 cm. Find the mean length, themean absolute error and the percentageerror in the measured value of thelength.[Ans: 3.13 cm, |
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Answer» Given: a1=3.11; a2=3.13; A3=3.14; A4=3.14 Mean LENGTH (am) = a1+a2+q3+a4/4 =3.11+3.13+3.14+3.14/4 = 12.52/4 (am) = 3.13cm. ABSOLUTE error: ∆a1=|am - a1 | = | 3.13-3.11 | = 0.02 ∆a2=|am -a2 | = | 3.13-3.13 | =0 ∆a3=|am -a3 | = | 3.13-3.14 | =0.01 ∆a4=|am -a4 | = | 3.13-3.14 | =0.01 Mean absolute error: ∆am=∆a1+∆a2+∆a3+∆a4/4 ∆am=0.02+0+0.01+0.01/4 ∆am=0.01cm % error = ∆am/am =0.01/3.13*100 =0.31% |
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